In: Chemistry
Determine the pH at equivalence when 12.0ml of 0.025mol/L benzoic acid (Ka=6.3x10-5) is titrated with 0.050mol/L NaOH.
find the volume of NaOH used to reach equivalence point
M(C6H5COOH)*V(C6H5COOH) =M(NaOH)*V(NaOH)
0.025 M *12.0 mL = 0.05M *V(NaOH)
V(NaOH) = 6 mL
we have:
Molarity of C6H5COOH = 0.025 M
Volume of C6H5COOH = 12 mL
Molarity of NaOH = 0.05 M
Volume of NaOH = 6 mL
mol of C6H5COOH = Molarity of C6H5COOH * Volume of C6H5COOH
mol of C6H5COOH = 0.025 M * 12 mL = 0.3 mmol
mol of NaOH = Molarity of NaOH * Volume of NaOH
mol of NaOH = 0.05 M * 6 mL = 0.3 mmol
We have:
mol of C6H5COOH = 0.3 mmol
mol of NaOH = 0.3 mmol
0.3 mmol of both will react to form C6H5COO- and H2O
C6H5COO- here is strong base
C6H5COO- formed = 0.3 mmol
Volume of Solution = 12 + 6 = 18 mL
Kb of C6H5COO- = Kw/Ka = 1*10^-14/6.3*10^-5 = 1.587*10^-10
concentration ofC6H5COO-,c = 0.3 mmol/18 mL = 0.0167M
C6H5COO- dissociates as
C6H5COO- + H2O -----> C6H5COOH + OH-
0.0167 0 0
0.0167-x x x
Kb = [C6H5COOH][OH-]/[C6H5COO-]
Kb = x*x/(c-x)
Assuming small x approximation, that is lets assume that x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((1.587*10^-10)*1.667*10^-2) = 1.627*10^-6
since c is much greater than x, our assumption is correct
so, x = 1.627*10^-6 M
[OH-] = x = 1.627*10^-6 M
we have below equation to be used:
pOH = -log [OH-]
= -log (1.627*10^-6)
= 5.79
we have below equation to be used:
PH = 14 - pOH
= 14 - 5.79
= 8.21
Answer: 8.21